A ball falls in the air from a very high height and hits the ground absolutely elastically. Find the acceleration of the ball immediately after the impact, considering the drag force acting on the ball from the air to be proportional to its velocity.
Decision:
A falling ball of mass $m$ is subject to the forces of gravity $mg$ and air resistance $F = -kv$, where $v$ is the velocity of the ball, $k$ is the coefficient of proportionality. Under the action of the resultant of these forces, the ball acquires acceleration
$a=\frac{mg+F}{m}=g-\frac{k}{m}v$ (1)
The acceleration is maximum at the initial moment and is equal to $g$. As the velocity increases, as follows from equation (1), the acceleration decreases and finally (the ball is falling from a very high height!) becomes almost zero. In this case, the velocity of the ball is constant and equal to
$v_{0}=\frac{m}{k}g$.
and there is a constant air resistance force acting on it.
$F_{0} = -kv_{0} = -mg$.
At absolutely elastic impact with the ground, the ball changes the direction of motion to the opposite direction and acquires velocity $v_{1} = v_{0}$. Consequently, the ball is subjected to a drag force from the air
$F_{1} = - kv_{1} = kv_{0} = mg$.
In addition, the force of gravity $mg$ acts on it. As a result, the ball is accelerated
$a = \frac{mg+F_{1}}{m} = 2g$.
Decision:
A falling ball of mass $m$ is subject to the forces of gravity $mg$ and air resistance $F = -kv$, where $v$ is the velocity of the ball, $k$ is the coefficient of proportionality. Under the action of the resultant of these forces, the ball acquires acceleration
$a=\frac{mg+F}{m}=g-\frac{k}{m}v$ (1)
The acceleration is maximum at the initial moment and is equal to $g$. As the velocity increases, as follows from equation (1), the acceleration decreases and finally (the ball is falling from a very high height!) becomes almost zero. In this case, the velocity of the ball is constant and equal to
$v_{0}=\frac{m}{k}g$.
and there is a constant air resistance force acting on it.
$F_{0} = -kv_{0} = -mg$.
At absolutely elastic impact with the ground, the ball changes the direction of motion to the opposite direction and acquires velocity $v_{1} = v_{0}$. Consequently, the ball is subjected to a drag force from the air
$F_{1} = - kv_{1} = kv_{0} = mg$.
In addition, the force of gravity $mg$ acts on it. As a result, the ball is accelerated
$a = \frac{mg+F_{1}}{m} = 2g$.